Question

我的任务是每天从本地下载站点下载McAfee病毒定义文件。文件名每天都在变化。下载网站的路径是“http://download.nai.com/products/licensed/superdat/english/intel/7612xdat.exe” 这个ID 7612每天都会因为不同的东西而改变，因此我无法对其进行硬编码。我必须找到一种方法来列出文件名，然后再提供它作为参数等。在stackoverflow网站上，我找到了某人的脚本，如果有人可以建议如何处理更改文件名，这将对我有用。这是我将要使用的脚本：

def download(url):
    """Copy the contents of a file from a given URL
    to a local file.
    """
    import urllib
    webFile = urllib.urlopen(url)
    localFile = open(url.split('/')[-1], 'w')
    localFile.write(webFile.read())
    webFile.close()
    localFile.close()

if __name__ == '__main__':
    import sys
    if len(sys.argv) == 2:
        try:
            download(sys.argv[1])
        except IOError:
            print 'Filename not found.'
    else:
        import os
        print 'usage: %s http://server.com/path/to/filename' % os.path.basename(sys.argv[0])

有人可以建议我吗？提前致谢

Answer 1

这是一个两步的过程。首先抓取索引页面寻找文件。第二，抓住最新版本并下载

import urllib
import lxml.html
import os
import shutil

# index page
pattern_files_url = "http://download.nai.com/products/licensed/superdat/english/intel"
# relative url references based here
pattern_files_base = '/'.join(pattern_files_url.split('/')[:-1])

# scrape the index page for latest file list
doc = lxml.html.parse(pattern_files_url)
pattern_files = [ref for ref in doc.xpath("//a/@href") if ref.endswith('xdat.exe')]
if pattern_files:
    pattern_files.sort()
    newest = pattern_files[-1]
    local_name = newest.split('/')[-1]
    # grab it if we don't already have it
    if not os.path.exists(local_name):
        url = pattern_files_base + '/' + newest
        print("downloading %s to %s" % (url, local_name))
        remote = urllib.urlopen(url)
        print dir(remote)
        with open(local_name, 'w') as local:
            shutil.copyfileobj(remote, local, length=65536)

每次更改文件名时，从Web下载文件

1 个答案: